MHB Cartesian product and symmetric difference

[fatineouahbi]

Let A,B,C be three sets . Prove Ax(BΔC)= (AxB) Δ (AxC)

I tried to start with this :

Let p be an arbitrary element of Ax(BΔC)
then p=(x,y) such that x ∈ A and y ∈ (BΔC)
x ∈ A and (y∈ B\C or y∈ C\B)
(x ∈ A and y ∈ B\C) or (x ∈ A and y ∈ C\B)

But I don't know how to continue or if I should even start with this .

 

[evinda]

Let A,B,C be three sets . Prove Ax(BΔC)= (AxB) Δ (AxC)

I tried to start with this :

Let p be an arbitrary element of Ax(BΔC)
then p=(x,y) such that x ∈ A and y ∈ (BΔC)
x ∈ A and (y∈ B\C or y∈ C\B)
(x ∈ A and y ∈ B\C) or (x ∈ A and y ∈ C\B)

But I don't know how to continue or if I should even start with this .

It is right so far.When does it hold that $p \in (A \times B) \triangle (A \times C)$ ?

 

[fatineouahbi]

It is right so far.When does it hold that $p \in (A \times B) \triangle (A \times C)$ ?

Hello :) Thank you , I think I may get it now ?

(x ∈ A and y ∈ B\C) or (x ∈ A and y ∈ C\B)
then p ∈ Ax(B\C) or p ∈ Ax(C\B)
then p ∈ (AxB) \ (AxC) or p ∈ (AxC) \ (AxB)
thus p ∈ (AxB) △ (AxC)
then Ax(BΔC) ‎⊂ (AxB) Δ (AxC)

Then I'll just try to go backwards maybe ?

 

[evinda]

Hello :) Thank you , I think I may get it now ?

(x ∈ A and y ∈ B\C) or (x ∈ A and y ∈ C\B)
then p ∈ Ax(B\C) or p ∈ Ax(C\B)
then p ∈ (AxB) \ (AxC) or p ∈ (AxC) \ (AxB)
thus p ∈ (AxB) △ (AxC)
then Ax(BΔC) ‎⊂ (AxB) Δ (AxC)

Well done, you are right :)

Then I'll just try to go backwards maybe ?

Yes, you pick an element in $(A \times B)\triangle (A \times C)$ and you need to show that it is also in $A \times (B \triangle C)$.

 

[fatineouahbi]

Well done, you are right :)

Yes, you pick an element in $(A \times B)\triangle (A \times C)$ and you need to show that it is also in $A \times (B \triangle C)$.

Thank you so much !